[Maths Class Notes] on Lagrange Interpolation Theorem Pdf for Exam

A polynomial is an algebraic expression that can have one or more terms. The meaning of “Poly” is “many” and “nominal” means “terms,” so, in other words, a polynomial is “many terms.” It can have constants, variables, and exponents. They all can be combined using mathematical operations like additions, subtraction, multiplication, and division except that a division by a variable is not allowed in a polynomial expression.

Example of a polynomial – 2xy² + 4x – 6 –> This polynomial has 3 terms which are 2xy², 4x, and 6

A polynomial can have one or more terms, but infinite numbers of terms are not allowed as well as the exponent can only be positive integers (0, 1, 2, …); hence 3xy-2 is not a polynomial.

Interpolation

Within a range of a discrete set of data points, interpolation is the method of finding new data points. It is a technique in which an estimate of a mathematical expression is found, taking any intermediate value for the independent variable. The main use of interpolation is to figure out what other data can exist outside of their collected data. Many professionals like photographers, scientists, mathematicians, or engineers use this method for their experiments. A common use is in the scaling of images when one interpolates the next position of a pixel based on the given positions of pixels in an image.

Lagrange Interpolation Theorem

This theorem is a means to construct a polynomial that goes through a desired set of points and takes certain values at arbitrary points. If a function f(x) is known at discrete points x_i, i = 0, 1, 2,… then this theorem gives the approximation formula for nth degree polynomials to the function f(x). More so, it gives a constructive proof of the theorem below:

For a point p(2,4), how do we represent it as a polynomial?

P(x) = 3

P(1) = 3

Similarly for a sequence of points, (2,3), (4,5) how can we find a polynomial to represent it?

P(x) = (x-4)/(2-4) * 3 + (x-2)/(4-2) * 5

P(2) = 3 and P(4) = 5

Going by the above examples, the general form of Lagrange Interpolation theorem can be gives as:

P(x) = (x – x₂) (x-x₃)/(x₁ – x₂) (x₁ – x₃) * y₁ + (x – x₁) (x-x₃)/(x₂ – x₁) (x₂ – x₃) * y₂ + (x – x₁) (x-x₂)/(x₃ – x₁) (x₃ – x₂) * y₃

Or P (x)= [sum_{i=1}^{3}] [P_{i}](x) [Y_{i}]                    

The Theorem – For n real values which are distinct: x₁, x₂, x₃, x₄,..x_n, and n real values (might not be distinct) y₁, y₂, y₃, y₄… y_n, a unique polynomial exists with real coefficients which satisfies the formula:

P(x_i) = y_i, i ϵ (1, 2,3, …, n) so that deg(P) < n

Proof of Lagrange Theorem

Let us consider an nth degree polynomial which is given by the below expression:

F(x) = A₀ (x-x₁) (x-x₂) (x-x₃)….(x-x_n) + A₁ (x-x₀) (x-x₂) (x-x₃)….(x-x_n) + ……..+ A_n (x-x₁) (x-x₂) (x-x₃)….(x-x_n-1)

Now we substitute values of our observations i.e. xi and we obtain the values of Ai:

So we put x = x₀ and we get A0 as below:

f(x₀) = y₀ = A₀ (x₀-x₁) (x₀-x₂) (x₀-x₂)….(x₀-x_n), the other terms become 0

Hence A₀ = y₀/(x₀-x₁) (x₀-x₂) (x₀-x₃)….(x₀-x_n)

Similarly for x1 we would get 

f(x₁) = y₁ =  = A₁ (x₁-x₀) (x₁-x₂) (x₁-x₃)….(x₁-x_n), the other terms become 0

Hence A₁ = y₁/(x₁-x₀) (x₁-x₂) (x₁-x₃)….(x₁-x_n)

In this way we can obtain all the values of As from A₂, A₃,… A_n

A_n = y_n/(x_n-x₀) (x_n-x₂) (x_n-x₃)….(x_n-x_n-1)

Now if we substitute all the values of As in the main function, we get Lagrange’s interpolation theorem:

F(x) =  y₀ * (x-x₁) (x-x₂) (x-x₃)….(x-x_n)/ (x₀-x₁) (x₀-x₂) (x₀-x3)….(x₀-x_n) + y₁ * (x-x₀) (x-x₂) (x-x₃)….(x-x_n)/ (x₁-x₀) (x₁-x₂) (x₁-x₃)….(x₁-x_n) + ……..+ y_n (x-x₁) (x-x₂) (x-x₃)….(x-x_n-1)/ (x_n–x₀) (x_n-x₂) (x_n-x₃)….(x_n-x_n-1)

Note: Lagrange’s theorem applies to both equally and non-equally spaced points. This means that all the values of x_s are not spaced equally.

Uses of Lagrange Interpolation Theorem – In science, a complicated function needs a lot of time and energy to be solved. This makes experiments difficult to run. In order to create a slightly less complex version of the original function, the interpolation method comes in use.

Conclusion 

The Lagrange theorem generalizes the well-established mathematical facts like a line is uniquely determined by 2 points, 3 points uniquely determine the graph of a quadratic polynomial, and so on. There is a caveat here i.e.; the points must have different x coordinates. The image enlargement technique uses principles of the Lagrange theorem in trying to describe the tendency of image data by using interpolation polynomials to estimate unknown data. This helps in image enlargement.

The theorem can be expressed in a mathematical formula as shown here: 

[P (x) = sum i = 1_{n}sum i = 1_{n}P_{i}P_{i}(x)]

Yi and it applies to all values of x, whether they are equally spaced or not.